Download - TRIAL ADDMATE SPM 2010 Pahang Paper 2 Answer
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8/8/2019 TRIAL ADDMATE SPM 2010 Pahang Paper 2 Answer
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SULIT 1 3472/23472/2@2010 HAKCIPTA PKPSM PAHANG
SCHEME ANSWER OF PAPER 2
No. PERATURAN PEMARKAHAN MARKAH MARKAHPENUH
1 x=2y-4 or y = 1
Substitute x or y in equation 2
y2
-3(y)*(2y-4) =6 or
or
1
5y2 -12y +6 =0 or 5x 2 +16x+8 =0Solve quadratic equations by using a formula orcompleting the squareUsing a formula
y =1
By completing the squareOr 1
y = 1.69, 0.71 10
x = -0.62, -2.58 1 52
6 4
1, 2
dy x
dxdy
xdx
(a) Equation of the tangenty-2 = 2(x-1)
y =2x
(b) approximate change in x
=8
1
1
1
1
1
or
1
16
3
i
ii
BA AC BC or BA AT BT or BT TC BC oranother valid method
2 AC y x
12
BT x y
1
1
1
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8/8/2019 TRIAL ADDMATE SPM 2010 Pahang Paper 2 Answer
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SULIT 3472/2
SULIT 2 3472/23472/2@2010 HAKCIPTA PKPSM PAHANG
AP y x
*12
BR k x y
*
1
BR h y x y
BR h y hx
Compare:1
1 ,2
k h k h
1 2,
3 3h k
1
1
2 7
4
(a) 1
1
5 210.5 10516.5
Q
Q
=35.5Interquartile range=35.5 -16.5
= 19
6
5
4
3
2
1
numberofstudents
marks23.5 50.540.530.520.510.50.5
Score mode = *23.5
1
or
1
1
1
3
1 7
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No.PERATURAN PEMARKAHAN MARKAH MARKA
HPENUH
5
2 2 22
2cos tan sec
2cos 1
cos2
LHS
x x
x
x
or any correct method
Shape of cosineAmplitude (maximum and minimum)ShiftedStraight line (gradient or y-intercept) for
3 cos 2x -1 = or 3 x
y
Number of solutions = 4
1
1
11111
1 8
6(a) used formula 2 1
2n
nS a n d
2
2
12
20 12 8
2 12 1 8 17602
12 4 4 1760 0
2 440 0
22 20 0
22@ 20
n
a
d
nS n
n n
n n
n n
n n
1
1
1n=20
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(b) Used 1nT a n d
108
108 12 1 8
13
nT
n
n
(c) 20
20
12 19 8
164
T
T
1
1
1
17
Section B
7 10log 2 x
0.30 0.48 0.60 0.70 0.78
10log y 0.54 0.68 0.77 0.86 0.92
Refer to the appendix 1
Minimum 1 point plotted correctly(Correct axes and uniform scales)5 *points are correct plottedLine of best fit (at least 3 points lies on the straight line andthe other must be balance)
2 n y k x
10log log log 2 y k n x 10 10 10log log 2 log y n x k
0.80m n 100.3 logc k 1.995k
1
1
1
11
1
11+1
1
10
8
(a)
20, 9
3
0,3 , 0, 3
x y
y
Q R
solve simultaneous equations
2 3 9
3 4 0
3@ 4
4 3 7
7, 4
y y
y y
y
x
P
1
1
1
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(b)
or
area of shaded region= 2
=
=
= 2 -0}+
=
(b) Volume of generated
V=
V= -
=
1
1
1
1
1
1
1 10
9
(a)
1 2
2
0,2
1
2
32
23
L
m m
m
y x
Used formula a point dividing a segment of a line
(b) (i)
2 3 1 2 0 10, 2 ,
3 3
6,6
x y
M
1
1
1
1
1
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8/8/2019 TRIAL ADDMATE SPM 2010 Pahang Paper 2 Answer
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(ii) use area of triangle
2
1 86
2 3
133
unit
(c) 2
2 2 2
2 2
2
46 6 4 0
3
4 83 3 12 64 0
3 9
WM WK
x y x y
x y x y
1
1
1
1
1 10
10(a) mean, (i)
8 6.40.8
p p
p
(ii)find
0 8 1 7 2 68 8 8
0 1 2
3
1 0.8 0.2 0.8 0.2 0.8 0.2
P X
C C C
or using 3 3 4 5 6 7 8P X P X P X P X P X P X P X
0.9988 or 0.99877
(b) use 5.3 0.0668P z
or4.8
0.1587P z
5.31.5
4.81
Solve simultaneous equation (until 1 variableleft)
0.2, 5
11
1
Or 1
1
1
Or 1
1
1
1
1+1
10
11(a)
Used trigonometric ratios3
tan5
30.96 0 0.54 rad
1
1
2
(b) Used Pythagoras Theorem to find OR2 23 5 5.831OR
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5.831 5 0.831PQ 5.831 0.54QR
3.149
Used trigonometric ratios to find PRO
tan PRO =
PRO = 59.04 or
1.031 r 3 1.031 3.093SP
Perimeter of shaded region3 3.093 0.831 3.149 10.073
10.07
1
1
Or 1
1
1
4
(c) Area of sector PRS
21 3 1.0312
4.6395
Area of sector QOR
21 5.831 0.542
9.1802
Area of triangle OPR
1 3 52
7.5
Area of shaded region9.182 7.5 4.6395
=6.32
1
Or 1
Or 1
1
1 4
Total marks 10Section C
12 (a) Used sine rulesin sin 325.7 4.9
sin 0.6164
38.06
Q
Q
PQR
1
1
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(b) Used cosine rule
2 2 24.5 2.7 5.7cos
2 4.5 2.7
cos 0.2037
101.76
S
S
PSR
(c) Used sine rule or cosine rule
*
4.9sin 109.94 sin32
8.692
PQ
PQ
(d) Area of triangle PQR
* follow through
Area of triangle PSR
1 4.5 2.7 sin101.752
Area of PQRS1 2
13.13 5.94819.078
A A
1
11
1
1
1
Or
1
1
1
10
13 20, 18, 200 x y z
(b)Ingredient 2007/ 2005 I w wI
A 150 36 5400B 120 72 8640C 200 108 21600D 130 144 18720
360 54360
(b) Used
54360
360
151
Iw I
w
I
I
3
1
1
1
PSR =101.75 0
19.08
or equivalent
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(c) 20052005
3322100 151
2200
Q
Q
(d)
130151
100196.3
1
1
1
1
10
14
(a)
. 80
1.
2.420 240 20000
21 12 1000
I x y
II y x
III x y
x y
(b) Draw correctly at least one straight line * Draw correctly all three straight linesR correctly shaded.Refer to appendix 2
(c) (i) 31 x 50(ii)
* *
* *
420 53 240 26
28500
420 5 240 75
20100
or
The maximum fees collected is 28500
1
1
1
111
1+1
1
110
15 (a) 0 10v
(b) 8 4a t
08a
(c) , 8-4t =0t = 2
2maxmax
10 8 2 2 2
18
v
v
1
1
1
1
1
1
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8/8/2019 TRIAL ADDMATE SPM 2010 Pahang Paper 2 Answer
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SULIT 10 3472/23472/2@2010 HAKCIPTA PKPSM PAHANG
(d)
2
2 2
10 8 2
210 4
3
s vdt t t dt
s t t t
At least two terms changes their index.
Stops instantaneously, 0v
210 8 2 0
5 1 0
5@ 1
t t
t t
t t
2 3210 5 4 5 53
266
3
s
s
1
1
1
1
10
t3
Or=
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8/8/2019 TRIAL ADDMATE SPM 2010 Pahang Paper 2 Answer
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SULIT 11 3472/23472/2@2010 HAKCIPTA PKPSM PAHANG
Appendix 1Graph 7(a)
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SULIT 12 3472/23472/2@2010 HAKCIPTA PKPSM PAHANG
Appendix 2
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8/8/2019 TRIAL ADDMATE SPM 2010 Pahang Paper 2 Answer
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SULIT 13 3472/23472/2@2010 HAKCIPTA PKPSM PAHANG
Graph 14(b)
END OF SCHEME ANSWER
SKEMA PEMARKAHAN TAMAT